Secondary 6 Design And Applied Technology Practice Paper: Design Process, Materials And Manufacturing, Structures And Mechanisms, Electricity And Electronics, Systems And Control, Evaluation And Sustainable Design

Secondary 6 Design and Applied Technology Practice Paper: Design Process, Materials and Manufacturing, Structures and Mechanisms, Electricity and Electronics, Systems and Control, Evaluation and Sustainable Design

2026-11-14

Design and Applied Technology – Secondary 6 – Practice Paper

S6 Design and Applied Technology paper, 24 parts on design process, ergonomics, material cost, tolerances, structures, electronics, control and carbon footprint.

Paper details　Secondary 6 Design and Applied Technology practice paper with 24 parts and marking points on the design process, ergonomics, material density and cost, tolerances, moulding output, beam and lever calculations, electronics, pneumatics, ADC resolution and carbon footprint.

Section A: Design Process and Creativity

Show the working and give a whole number or the rounding asked.

#	Question	Answer and marking notes
1	A design project lasts 20 weeks: research 30 per cent, idea development 20 per cent, detail design and making 35 per cent, evaluation 15 per cent. Find the weeks allowed for each stage.	Answer: research 20 x 0.30 = 6 weeks, idea development 4 weeks, detail design and making 7 weeks, evaluation 3 weeks, total 20 weeks.
2	A seated worktop height is set from popliteal height plus seated elbow height above the seat. User A measures 400 mm and 240 mm, user B 450 mm and 260 mm. Find each required height and the adjustable range.	Answer: A needs 400 + 240 = 640 mm and B needs 450 + 260 = 710 mm, so the range is 640 to 710 mm, covering the 5th to the 95th percentile.
3	Six concept models are needed, each taking 3.5 hours. Two designers work in parallel for 7 hours a day. Find the number of working days needed.	Answer: each designer makes 3 models, taking 3 x 3.5 = 10.5 hours; 10.5 / 7 = 1.5 days, so 2 working days are needed.
4	Unit material cost: ABS 0.35 kg at 90 dollars per kg, aluminium 0.12 kg at 180 dollars per kg, fasteners 6 dollars and packaging 4 dollars. Find the unit cost, then the amount and percentage to cut to meet a 48 dollar limit, to 1 decimal place.	Answer: 31.5 + 21.6 + 6 + 4 = 63.1 dollars; the cut needed is 63.1 – 48 = 15.1 dollars, which is 15.1 / 63.1 = 23.9 per cent.

Section B: Materials and Manufacturing Processes

Show the working and state the units.

#	Question	Answer and marking notes
5	An aluminium sheet measures 1200 x 600 x 1.5 mm and aluminium has a density of 2700 kg per cubic metre. Find its mass in kg to 2 decimal places.	Answer: the volume is 1.2 x 0.6 x 0.0015 = 0.00108 cubic metres, so the mass is 0.00108 x 2700 = 2.92 kg.
6	Each sheet has an area of 720000 square millimetres and yields 12 blanks of 45000 square millimetres each. Find the material utilisation to 1 decimal place, then the material cost per part for 100 parts when a sheet costs 85 dollars, to 2 decimal places.	Answer: utilisation is 12 x 45000 / 720000 = 75.0 per cent; 100 parts need 9 sheets costing 9 x 85 = 765 dollars, so each part costs 765 / 100 = 7.65 dollars.
7	A hole is specified at 12.00 mm with limits plus 0.05 and zero; a shaft is 11.95 mm with a tolerance of plus or minus 0.02 mm. Find the minimum and maximum clearance and state whether the fit is clearance or interference.	Answer: the hole runs from 12.00 to 12.05 mm and the shaft from 11.93 to 11.97 mm; the minimum clearance is 12.00 – 11.97 = 0.03 mm and the maximum is 12.05 – 11.93 = 0.12 mm, all positive, so it is a clearance fit.
8	An injection moulding machine has 4 cavities and a 28 second cycle. Find the output per hour to 1 decimal place; for one 8 hour shift counted in complete cycles with a 3 per cent reject rate, find the good parts.	Answer: 3600 / 28 x 4 = 514.3 parts per hour; in 8 hours there are 1028 complete cycles giving 4112 parts, and 4112 x 0.97 = 3988 good parts.

Section C: Structures and Mechanisms

Show the working and give forces in newtons.

#	Question	Answer and marking notes
9	A simply supported beam is 3.6 m long with a central point load of 480 newtons. Find the reactions at the supports and the maximum bending moment in newton metres.	Answer: by symmetry each reaction is 480 / 2 = 240 newtons; the maximum moment is load times span divided by 4, 480 x 3.6 / 4 = 432 newton metres.
10	A lever carries a 180 newton load 0.25 m from the fulcrum, with the effort applied 1.0 m from the fulcrum on the other side. Find the effort needed and the mechanical advantage.	Answer: 180 x 0.25 = effort x 1.0, so the effort is 45 newtons and the mechanical advantage is 180 / 45 = 4.
11	A driver gear of 60 teeth runs at 900 revolutions per minute and drives a 20 tooth gear. Find the driven speed and the gear ratio.	Answer: the speed ratio is 60 / 20 = 3, so the driven gear turns at 900 x 3 = 2700 revolutions per minute, a ratio of 3 to 1 with a speed increase.
12	A pulley system supports a 240 newton load on 4 rope sections with an efficiency of 80 per cent. Find the actual effort and the input and useful work in joules when the load rises 0.5 m.	Answer: the effort is 240 / (4 x 0.8) = 75 newtons; the useful work is 240 x 0.5 = 120 joules and the input work is 75 x 4 x 0.5 = 150 joules, giving 120 / 150 = 80 per cent.

Section D: Electricity and Electronics

Show the working; currents may be given in amperes or milliamperes.

#	Question	Answer and marking notes
13	A 240 ohm resistor is connected across a 12 volt supply. Find the current in milliamperes and the power in watts.	Answer: the current is 12 / 240 = 0.05 amperes, that is 50 milliamperes, and the power is 12 x 0.05 = 0.6 watts.
14	Resistors of 220 ohm and 330 ohm are connected first in series and then in parallel. Find both total resistances in ohms.	Answer: in series 220 + 330 = 550 ohms; in parallel 220 x 330 / 550 = 132 ohms.
15	A light emitting diode has a forward voltage of 2.1 volts at 20 milliamperes and is supplied from 9 volts. Find the series resistor in ohms; if 330 ohms is used instead, find the actual current in milliamperes to 1 decimal place.	Answer: the resistor is (9 – 2.1) / 0.02 = 345 ohms; with 330 ohms the current is 6.9 / 330 = 20.9 milliamperes.
16	A battery is rated 2000 milliampere hours at a nominal 3.7 volts and the circuit draws 25 milliamperes continuously. Find the running time in hours and the stored energy in watt hours, then check the time from the power.	Answer: the running time is 2000 / 25 = 80 hours; the energy is 3.7 x 2.0 = 7.4 watt hours; the power is 3.7 x 0.025 = 0.0925 watts and 7.4 / 0.0925 = 80 hours, which agrees.

Section E: Systems and Control

Show the working and give the rounding asked.

#	Question	Answer and marking notes
17	A cylinder has a 32 mm bore at a supply pressure of 0.6 megapascals, that is 0.6 newtons per square millimetre. Find the piston force in newtons to 1 decimal place.	Answer: the area is pi times 16 squared = 804.2 square millimetres, so the force is 0.6 x 804.2 = 482.5 newtons.
18	The same cylinder has a 200 mm stroke completed in 1.5 seconds. Find the average speed in millimetres per second and in metres per minute to 1 decimal place.	Answer: 200 / 1.5 = 133.3 millimetres per second, which is 133.3 x 60 / 1000 = 8.0 metres per minute.
19	A 10 bit analogue to digital converter has a 5 volt reference, and a sensor gives 0 to 5 volts for 0 to 100 degrees. Find the resolution in millivolts to 1 decimal place and the temperature at a reading of 512.	Answer: the resolution is 5 / 1024 = 4.9 millivolts; a reading of 512 gives 5 x 512 / 1024 = 2.5 volts, that is 50 degrees.
20	A motor runs at 3000 revolutions per minute through a 50 to 1 reduction gearbox driving a wheel of 60 mm diameter. Find the wheel speed in revolutions per minute and the travel speed in metres per minute to 1 decimal place.	Answer: the wheel turns at 3000 / 50 = 60 revolutions per minute; its circumference is 0.06 x pi = 0.188 m, so the speed is 60 x 0.188 = 11.3 metres per minute.

Section F: Product Evaluation and Sustainable Design

Show the working and give the rounding asked.

#	Question	Answer and marking notes
21	A 9 watt LED lamp replaces a 50 watt halogen lamp, used 4 hours a day for 365 days a year. Find the annual energy saved in kilowatt hours to 2 decimal places.	Answer: (50 – 9) x 4 x 365 = 59860 watt hours, that is 59.86 kilowatt hours.
22	A product has a mass of 1.25 kg of which 68 per cent is recyclable. Find the recyclable mass per unit in kg and the total for 1000 units.	Answer: 1.25 x 0.68 = 0.85 kg per unit, so 1000 units give 0.85 x 1000 = 850 kg.
23	A lamp costs 180 dollars and saves 71.83 dollars of electricity a year. Find the payback period in years to 1 decimal place.	Answer: 180 / 71.83 = 2.5 years.
24	The annual saving is 59.86 kilowatt hours and the grid factor is 0.55 kilograms of carbon dioxide equivalent per kilowatt hour. Find the annual carbon reduction and the 5 year total in kg to 1 decimal place.	Answer: 59.86 x 0.55 = 32.9 kg a year, and 32.92 x 5 = 164.6 kg over 5 years.

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